I come across the interesting egg drop problems and this two-egg variation.
Briefly, the problem is: Given a building with n floors and two eggs, we want to find the critical floor f, where the egg breaks if dropped from floor $≥$ f, while stays intact and can be reused if from floor $<$ f. And the goal is to find the minimum times of attempts to find the f.
The solution is to try integer $ x \in [0, \infty) $ such that:
If the first egg breaks on the first drop, i.e., floor $x$, we need up to $x - 1$ more trials with the second egg: total $x$ trials.
If it breaks on the second drop, i.e., floor $x + (x - 1)$, we've already used 2 trials, and need up to $x - 2$ more: total $x$ trials.
...
If it doesn't break until the last drop, the number of trials is x.
All in all, find $x$ such that $x + ( x - 1 ) + ( x - 2 ) + ... + 1 >= n$. Surely one can simply solve $\frac{x(x + 1)}{2} \geq n$.
Intuitively, I understand the logic behind the solution is to balanced the risk across all scenarios so as to minimize the overall cost. Just like the way we apply bisection if we have infinite eggs, the tradeoff between gain of success vs risk of failure is well balanced fifty-fifty.
I, however, failed to translate the intuition into a mathematical proof. Could you kindly teach how to properly conduct a formal proof for it?
